# Tag Info

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Yeaaahh, I got it! :D Great thanks to David C. Ullrich!!! Counterexample Given the Hilbert space $\mathbb{C}^4$. Regard the matrices: $$N:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad N':=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ Then they are ...

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Thank you for the comments. I was able to use them to provide an answer to When does $\sum_{i=1}^{\infty} X_i$ exist for random sequences $\{X_i\}_{i=1}^{\infty}$?. I'll restate the result from there as it provides an example: Consider the measure space $L^2([0,1])$ with uniform lebesgue measure. Define the functions (Haar functions) $f_{2^i + k}$ by ...

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If one of the members of the product was NOT invertible,it would map a sequence of vectors, that doesn't converge to zero, to a sequence converging to zero.But then the product would also do that, making the product non-invertible.

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With a clear domain definition, and $\|\frac{1}{\Delta t}\{U_{\alpha}(\Delta t)-1_\alpha\}\varphi_{\alpha}-H_{\alpha}\varphi_{\alpha}\|=\|\frac{1}{\Delta t}\int_{0}^{\Delta t}(U_{\alpha}(t)-1_\alpha)H_{\alpha}\varphi_\alpha dt\|$, can you now better establish convergence?

3

Another Banach-algebra structure on the Hilbert space $\mathsf{hs}(H)$ of Hilbert-Schmidt operators on a Hilbert space $H$ is just operator multiplication (composition). There is a natural involution on this algebra but it does not make it a C*-algebra.

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Left Shift Denote for shorthand: $$1_0(0,x_1,\ldots):=(0,x_1,\ldots)$$ For the modulus: $$L^*L=RL=1_0\implies|L|=1_0$$ For the argument: $$L=U|L|\implies U=1_0,1,\ldots$$ So it admits one. Right Shift For the modulus: $$R^*R=LR=1\implies|R|=1$$ For the argument: $$R=U|R|\implies U=R$$ So it admits none. Reference For much more details: Polar ...

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For your second question, the answer is always yes. More generally, given any Banach algebra $A$, let $\tilde{A}=A\oplus \mathbb{C}$ be its unitization (with norm $\|(a,z)\|=\|a\|+|z|$). Given $a\in A$, let $L_a\in B(\tilde{A})$ be left multiplication by $a$. Then $a\mapsto L_a$ is an isometric isomorphism from $A$ to a subalgebra of $B(\tilde{A})$. ...

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As noted in my comment, $E$ has to be a Hilbert space. To see this, fix a linear functional $\varphi \in E'$ with $\Vert \varphi \Vert = 1$. For $z \in E$, define $$A_z : E \to E, x\mapsto \varphi(x) \cdot z.$$ It is not hard to see that $E \to B(E), z \mapsto A_z$ is linear and isometric. Thus, if $B(E) = H$ is a Hilbert space, we see that $E$ is ...

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Let $J$ be an index for the cardinality of an orthonormal basis of $H$. Then $H$ is isometrically isomorphic to $\ell^2 (J)$, so it is enough to discuss the problem on this latter space. Define the product $fg$ pointwise, i.e. $fg (j):=f (j)g (j)$. The question is whether this product stays in $\ell^2$, and whether the norm is submultiplicative. We ...

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I know a very special case, and I guess it is extensible. Suppose that $H$ is Hilbert space such that $H\cong (H_1\hat{\otimes}(H_2)^*)^*$, where $\hat{\otimes}$ is projective tensor product, and $H_1,H_2$ is some Hilbert spaces. For any Hilbert space $\mathcal{H}$, always $\mathcal{H}^{**}=\mathcal{H}$. Also for two Banach space $E,F$ always ...

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Let $\mathcal S$ be an orthonormal basis of eigenvectors for $|A|$ (this exists, since $|A|$ is trace-class and thus compact). Let $\mathcal T$ be any other orthonormal basis. For any $\sigma\in\mathcal S$, we have $$|A|\sigma=s_\sigma(A)\,\sigma$$ and $\sum_\sigma s_\sigma(A)=\sum_\sigma\langle |A|\sigma,\sigma\rangle<\infty$. Then, for any $B\in ... 1 Yes. The functional calculus preserves approximation by polynomials. So, from $$T^n=\begin{bmatrix}A^n&0\\0&B^n\end{bmatrix},$$ you get that $$p(T)=\begin{bmatrix}p(A)&0\\0&p(B)\end{bmatrix}$$ for any polynomial$p$. Now using a sequence of polynomials that converges uniformly to$f, you get that $$... 3 Left/right shift operators are the standard examples, but I personally think that the multiplication operator is the easiest way to see that there may be something else in the spectrum besides eigenvalues. Consider a multiplication operator A_c on \ell^\infty (c\in\ell^\infty)$$ (A_c x)_n=c_n x_n. The inverse if exists is clearly a multiplication ... 7 If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's x \ne y such that (T - \lambda I)(x) = (T - \lambda I)(y). So (T - \lambda I)(x - y) ... 6 For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators L, R:\ell^2 \to \ell^2, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map L ... 3 T-\lambda I being non-invertible does not imply there is a non-zero x with (T-\lambda I)x=0. That is true when H is finite-dimensional, but not necessarily when H is infinite-dimensional. The classic counterexample is the right-shift operator R:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}). Take a look at the Wikipedia article on the notion of ... 1 Although 0 \le A \le B implies \overline{{\cal R}(A)} \subseteq \overline{{\cal R}(B)}, it's not true without the closures. For a counterexample, take L^2[0,1]. Let A be multiplication by x (i.e. A f(x) = x f(x)), and B = A + u u^* where u(t) = t^{1/4} and u^* is the corresponding linear functional, i.e. B f(x) = A f(x) + u^*(f) u(x) = ... 1 This is always true, as\Omega \setminus (\Omega_1\cup \Omega_2)$can only contain the boundaries of$\Omega_1$and$\Omega_2$, which have zero measure due to the regularity assumptions on the domains. 0 Weak convergence in$H=H^1_0$implies strong convergence in$L^2$because$H^1_0$is compactly embedded in$L^2$, see: https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem 1 The implication$0\leq A\leq B\implies \mathcal RA\subset\mathcal RB$can be proven as follows. From$0\leq A\leq B$, we easily see that if$Bx=0$, then $$0\leq\langle Ax,x\rangle\leq\langle Bx,x\rangle=0,$$ so$A^{1/2}x=0$, and thus$Ax=0$. In other words,$\ker B\subset\ker A$. Then $$\overline{\mathcal RA}=(\ker A)^\perp\subset(\ker ... 0 No, no, and no. No, E is not even a Hilbert space! If we say d\mu = w dt then the completion of E is H=L^2(\mu), which is a Hilbert space. No, H is not a reproducing-kernel Hilbert space. It can't be, since it's not even a space of functions! (More formally, the map f\mapsto f(0) is not even defined on H.) It still could be that there is a ... 1 Presumably, you mean for H to be a vector space over \Bbb C. Note that \operatorname{Herm}(H^k) is a vector space over \Bbb R (as opposed to \Bbb C), and that \dim \operatorname{Herm}(H^k) = k^2. From there, we can simply apply your earlier reasoning to find that$$ \mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H^k)^{\otimes N})) = \binom{N + k^2 - ... 1 If$\alpha\neq 1$, then it fails for$x=y$, since the right hand side is$0$, and the left hand side is positive. 0 Existence For complex sums: $$\sum_\sigma\langle A\sigma,\sigma\rangle\in\mathbb{C}\iff\sum_\sigma|\langle A\sigma,\sigma\rangle|<\infty$$ By polar decomposition: $$|\langle A\sigma,\sigma\rangle|\leq\||A|^{1/2}\sigma\|\cdot\||A|^{1/2}J^*\sigma\|$$ For partial isometries: $$\tau^{(\prime)}:=J^*\sigma^{(\prime)}:\quad\sigma\perp\sigma'\implies ... 0 The assertion is false as: Given the Hilbert space \ell^2(\mathbb{N}). Consider the right shift:$$A:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A:=R$$Then it has finite trace:$$\sum_n\langle A\delta_n,\delta_n\rangle=\sum_n0=0$$But it is not trace class:$$\sum_n\langle|A|\delta_n,\delta_n\rangle=\sum_n1=\infty$$Concluding counterexample. 0 Meanwhile I got it... Limit Projection By orthogonality one has:$$\varphi\in\mathcal{R}^\perp\implies\varphi\in\mathcal{R}_\lambda^\perp\quad(\lambda\in\Lambda)$$So one obtains:$$\|P_\lambda\varphi-0\|=\|0-0\|=0\stackrel{\lambda}{\to}0$$By monotony one has: ... 1 The solution sets generally won't be the same. Try the example$$ A = \pmatrix{0&1\\1&0}, \\ h_1 = 0, \quad h_2 = 1 $$In this case, the first equation gives us$$ \Re\{\psi_1\psi_2\} = 0 $$Whereas the second gives us$$ \Im\{\psi_1\psi_2\} = 0 $$The two equations have the same solution if and only if A = 0. Otherwise, suppose that A_{ij} ... 1 Seems to me that the definition given for H^s can't be right for s<0; it "must" be that H^s is actually the space of tempered distributions f such that \hat f\in L^2(\mu), where d\mu(\xi)=(1+|\xi|^2)^{s/2}\,d\xi. Assuming so, this is easy: Say (f_n) is Cauchy in H^s. Then (\hat f_n) is Cauchy in L^2(\mu). So \hat f_n\to g in ... 1 The thing about C(\Omega) works because \mathbb R is complete: If (f_n) is a Cauchy sequence in C(\Omega) then the definition of the metric shows that for every x\in\Omega the sequence (f_n(x)) is a Cauchy sequence of reals. Since \mathbb R is complete there exists a real number y such that f_n(x)\to y; we define f(x)=y and proceed. The ... 3 You know the u_m are converging to a function, because \mathbb{R} is complete. For every x \in \Omega, you have$$| u_n(x) - u_m(x) | \leq \| u_n - u_m\|_\infty \to 0$$So u_n(x) is a cauchy sequence, and by completness of \mathbb{R}, converge. The function u is then defined by :$$u(x) = \lim_{n\to \infty} u_n (x)$$For your second ... 4 Let x \in \mathcal{H} with \lVert x\rVert = 1. Show that x cannot be in the orthogonal complement of \operatorname{span} \{ f_n : n \in \mathbb{N}\}. Since \{ e_n : n \in \mathbb{N}\} is a Hilbert basis of \mathcal{H}, we can write$$x = \sum_{n = 0}^\infty c_n\cdot e_n.$$Consider now$$y = \sum_{n = 0}^\infty c_n \cdot f_n.$$Then we have ... 1 Consider the pushforward:$$E_\eta:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E_\eta(A):=E(\eta^{-1}A)$$Their domain agree as:*$$\int|\vartheta\circ\eta|^2\mathrm{d}\nu_\varphi(\lambda)=\int|\vartheta|^2\mathrm{d}\nu^\eta_\varphi$$Denote for shorthand: ... 1 In M_2(\mathbb C), the normal operators are precisely the unitarily diagonalizable ones. So, any non-selfadjoint normal operator is, as TrialAndError mentioned, of the form$$ N=U^*\,\begin{bmatrix}\lambda&0\\0&\mu\end{bmatrix}\,U $$with at least one of \lambda,\mu not real (if both were real, N=N^*). The general 2\times 2 unitary is of ... 1 In general, they are not weakly closed (not even strongly closed). To see this, consider the Hilbert space \ell^2([0,1]) and the operator N=M_{\rm id}. Then the (bounded) Borel functional calculus will yield all operators M_f (multiplication with f, where f is Borel measurable and bounded. But now consider the set A:=\{I\subset [0,1]\,\mid\,I ... 2 This is a (probably icomplete) hierarchy of vector spaces from the point of view of functional analysis. Vector spaces: algebraic structure (addition and multiplication by scalars) Topological vector space: vector space with a topology such that addition and multiplication by scalars are continuous. Locally convex topological vector spaces: TVS in which ... 1 On H=\mathbb C^2, for each \lambda\in \mathbb C let P_\lambda be the projection onto the span of \{(1,\lambda)\}. Then P_\lambda\to P_0 as \lambda\to 0, but for \lambda\neq 0, P_0\land P_\lambda =0, while P_0\land P_0=P_0\neq 0. Also, for \lambda\neq 0, P_0\lor P_\lambda = I, while P_0\lor P_0 = P_0\neq I. 0 Meanwhile I got it... Define the function:$$\varphi\in\mathcal{D}H_0^2:\quad\omega(t):=U(t)^*JU_0(t)\varphi$$Its derivatives are:$$\omega'(t)=U(t)i\{HJ-JH_0\}U_0(t)\varphi\omega''(t)=U(t)i^2\{H^2J-2HJH_0+JH_0^2\}U_0(t)\varphi$$By Pettis' criterion:* ... 0 Suppose a sequence (h_n+Th_n) in \hat{H} converges to some x, i.e., h_n+Th_n\to x. Applying T and using the fact that T^2=id, we obtain$$Tx=\lim T(h_n+Th_n)=\lim Th_n+h_n=x,$$so x=\frac{x+Tx}{2}=\frac{x}{2}+T\frac{x}{2}\in\hat{H}, which shows that \hat{H} is closed. 1 The result is indeed true. We have S = TT^* - T^*T. As you said, we may conclude that \langle x, Sx \rangle = 0 for all x \in X, which in this case is a real vector space. We may use this information conclude that S is skew-adjoint, that is, S^* = -S. We then additionally note that$$ S^* = (TT^*)^* - (T^*T)^* = TT^* - T^*T = S $$So that S ... 0 Meanwhile I got it... Projection They have domain:$$\mathcal{D}(P^*P)=\mathcal{D}(P^2)=\mathcal{D}P$$So it must be bounded:$$\|P\varphi\|^2=\langle P^*P\varphi,\varphi\rangle=\langle P\varphi,\varphi\rangle\leq\|P\varphi\|\cdot\|\varphi\|$$Also dense and closed:$$P=P^*=P^{**}\implies P=\overline{P}\quad(\overline{\mathcal{D}P}=\mathcal{H})$$By ... 1 For the "only if" direction, consider the homogenous system Ax = 0. Then the i-th row is$$\sum_{j=1}^n \langle h_j, h_i \rangle x_j = 0$$By linearity in the first argument it is$$\langle \sum_{j=1}^n x_j h_j, h_i \rangle = 0$$Therefore \sum_{j=1}^n x_j h_j \in \text{span}(h_1,h_2,\ldots,h_n)^\perp. On the other hand \sum_{j=1}^n x_j h_j is a ... 0 Note that one has:$$(\mathcal{N}\Omega)^\perp=\overline{\mathcal{R}\Omega^*}=\overline{\mathcal{R}|\Omega|}$$Denote embeddings: ... 0 Denote for shorthand:$$\Omega':=\Omega^*\Omega\in\mathcal{B}(\mathcal{H}):\quad |\Omega|=\lim_n\Omega'_n$$Regard spectral measures:$$H=\int\lambda\mathrm{d}E(\lambda)\quad H_0=\int\lambda\mathrm{d}E_0(\lambda)$$By a previous thread:*$$\Omega E_0(A)=E(A)\Omega\implies\Omega'E_0(A)=E_0(A)\Omega'\implies(\Omega')^n E_0=E_0(A)(\Omega')^n$$So one arrives ... 2 We must assume {\mathcal D}A is dense, otherwise A^* is not well defined. If x \in {\mathcal D}A and Ax = 0 then of course A^*A = 0. For the converse, note that y \in {\mathcal D} A^* with A^* y = z iff for every w \in {\mathcal D} A, \langle A w, y\rangle = \langle w, z\rangle. Now x \in {\mathcal N} A^* A iff x \in {\mathcal D} A ... 1 On the one hand:$$\varphi\in\mathcal{N}A\implies A\varphi=0\in\mathcal{D}A^*\implies\varphi\in\mathcal{N}(A^*A)$$On the other hand:$$\varphi\in\mathcal{N}(A^*A)\implies\|A\varphi\|^2=\langle A^*A\varphi,\varphi\rangle=0\implies\varphi\in\mathcal{N}A$$Concluding the assertion. 0 Rudin restricts to normals as... Definition Given a Hilbert space \mathcal{H}. Consider selfadjoint operators:$$H:\mathcal{D}H\subseteq\mathcal{H}\to\mathcal{H}:\quad H=H^*$$Define positivity by:*$$H\geq0:\iff\sigma(H)\geq0$$*(They may be unbounded!) Positive Operators Given a Hilbert space \mathcal{H}. Consider positive operators: ... 3 If \{e_n\} is an orthonormal basis for a Hilbert space H, then$$H = \left\{ \sum_{n=1}^\infty \alpha_n e_n : \sum_{n=1}^\infty |\alpha_n|^2 < \infty \right\}.$$For f=\sum \alpha_n e_n \in H we formally define the operation:$$Af = \sum_{n=1}^\infty \lambda_n \alpha_n e_n.$$This operator is linear, and it is a well defined map from H \to H ... 1 Meanwhile I got it: False! Given the Hilbert space \ell^2(\mathbb{N}). Consider right shifts:$$R_n:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad R_n:=R^n$$They are isometric:$$\|R^n\varphi\|=\|\varphi\|\quad(\varphi\in\ell^2)$$But only weakly$$R_n\rightharpoonup0\quad R_n\not\to0$$Concluding counterexample. See also: Weak vs. Strong 0 By unitarity one has:$$\|\Omega\varphi\|=\lim_{t\to\infty}\|JU_0(t)\varphi\|$$It can be written as:$$\|JU_0(t)\varphi\|^2=\langle\{J^*J-1\}U_0(t)\varphi,U_0(t)\varphi\rangle+\|\varphi\|^2$$But it is bounded:$$\|U_0(t)\varphi\|_{t\in\mathbb{R}}\equiv\|\varphi\|_{t\in\mathbb{R}}<\infty$$So one obtains: ... 1 Without loss of generality we can assume that$z=0$(otherwise apply the proof to the operator$A-z$instead). The sequence$\{f_n\}$is Weyl, i.e.$\|f_n\|=1$and$\|Af_n\|\to 0$when$n\to+\infty$. The vector$(f_n,Af_n)$belongs to the graph$\Gamma(A)$. Since$A=\bar A_0$we know that$\Gamma(A)=\bar\Gamma(A_0)$, so we can approximate$(f_n,Af_n)\$ ...

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